Problem: $f(x,y) = \sin(x) - 3\cos(y)$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $3\sin(y)$ (Choice B) B $\cos(x)$ (Choice C) C $-\cos(x)$ (Choice D) D $-3\sin(y)$
Solution: We want to find $\dfrac{\partial f}{\partial y}$, which is the partial derivative of $f$ with respect to $y$. When we take a partial derivative with respect to $y$, we treat $x$ as if it were a constant. Let's break $f(x, y)$ down term by term. $\begin{aligned} &\dfrac{\partial}{\partial y} \left[ \sin(x) \right] = 0 \\ \\ &\dfrac{\partial}{\partial y} \left[ -3\cos(y) \right] = 3\sin(y) \end{aligned}$ Adding the terms back together, we get the partial derivative. In conclusion: $\dfrac{\partial f}{\partial y} = 0 + 3\sin(y) = 3\sin(y)$